Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $k = \dfrac{8r + 40}{4r - 4} \times \dfrac{3r - 3}{r^2 - 25} $
Explanation: First factor the quadratic. $k = \dfrac{8r + 40}{4r - 4} \times \dfrac{3r - 3}{(r + 5)(r - 5)} $ Then factor out any other terms. $k = \dfrac{8(r + 5)}{4(r - 1)} \times \dfrac{3(r - 1)}{(r + 5)(r - 5)} $ Then multiply the two numerators and multiply the two denominators. $k = \dfrac{ 8(r + 5) \times 3(r - 1) } { 4(r - 1) \times (r + 5)(r - 5) } $ $k = \dfrac{ 24(r + 5)(r - 1)}{ 4(r - 1)(r + 5)(r - 5)} $ Notice that $(r - 1)$ and $(r + 5)$ appear in both the numerator and denominator so we can cancel them. $k = \dfrac{ 24\cancel{(r + 5)}(r - 1)}{ 4(r - 1)\cancel{(r + 5)}(r - 5)} $ We are dividing by $r + 5$ , so $r + 5 \neq 0$ Therefore, $r \neq -5$ $k = \dfrac{ 24\cancel{(r + 5)}\cancel{(r - 1)}}{ 4\cancel{(r - 1)}\cancel{(r + 5)}(r - 5)} $ We are dividing by $r - 1$ , so $r - 1 \neq 0$ Therefore, $r \neq 1$ $k = \dfrac{24}{4(r - 5)} $ $k = \dfrac{6}{r - 5} ; \space r \neq -5 ; \space r \neq 1 $